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Question Number 122298 by mnjuly1970 last updated on 15/Nov/20

$$...nicecalculus...$$$$provethat:$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\left\{\frac{\zeta (2n+1)-1}{n+1}\right\}=-\gamma +ln\left(2\right)✓$$$$..m.n.1970..$$

Answered by mindispower last updated on 17/Nov/20

$$\zeta \left(s\right)-1={\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{\mathrm{\Gamma }\left(s\right)\left(e^x(e^x-1)\right)}dx$$$$\Rightarrow \frac{\zeta (2n+1)-1}{n+1}={\int }_0^{\mathrm{\infty }}\frac{x^{2n}dx}{\mathrm{\Gamma }(2n+1)e^x(e^x-1)}$$$$\sum _{n⩾1}\frac{\zeta (2n+1)-1}{n+1}={\int }_0^{\mathrm{\infty }}\frac{x^{2n}dx}{(n+1)\mathrm{\Gamma }(2n+1)e^x(e^x-1)}..I$$$$\mathrm{\Sigma }\frac{x^{2n}}{2n!(n+1)}=\frac{1}{x^2}\sum _{n⩾1}\frac{x^{2n+2}}{2n!(n+1)}=g\left(x\right)$$$$\sum _{n⩾1}\frac{x^{2n+2}}{2n!(n+1)}=f\left(x\right)\Rightarrow f^{\prime }\left(x\right)=x\sum _{n⩾1}\frac{2x^{2n}}{(2n!)}$$$$=2x(ch\left(x\right)-1)$$$$=2xsh\left(x\right)-2ch\left(x\right)-x^2+2$$$$f\left(x\right)=2xsh\left(x\right)-x^2-2ch\left(x\right)+2$$$$g\left(x\right)=\frac{2sh\left(x\right)}{x}-1-\frac{2ch\left(x\right)}{x^2}+\frac{2}{x^2}$$$${\int }_0^{\mathrm{\infty }}\frac{2xsh\left(x\right)-x^2-2ch\left(x\right)+2}{x^2{e}^x(e^x-1)}dx=S$$$$={\int }_0^{\mathrm{\infty }}\frac{{xe}^x-{xe}^{-x}-x^2-e^x-e^{-x}+2}{x^2{e}^x(e^x-1)}dx$$$$$$$$={\int }_0^{\mathrm{\infty }}\frac{{xe}^{-x}-{xe}^{-3x}-x^2{e}^{-2x}-e^{-x}-e^{-3x}+2e^{-2x}}{x^2(1-e^{-x})}dx..E$$$$\mathrm{\Psi }\left(z\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{t}-\frac{e^{-zt}}{1-e^{-t}}dt$$$${\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{t}-\frac{e^{-t}}{1-e^{-t}}dt=\mathrm{\Psi }\left(1\right)$$$$={\int }_0^{\mathrm{\infty }}\frac{e^{-t}-{te}^{-t}-e^{-2t}}{t(1-e^{-t})}dt..nice$$$$E={\int }_0^{\mathrm{\infty }}\frac{{te}^{-t}-t^2{e}^{-t}-{te}^{-2t}+t^2{e}^{-t}+{te}^{-2t}-{te}^{-3t}-t^2{e}^{-2t}-e^{-t}-e^{-3t}+2e^{-2t}}{t^2(1-e^{-t})}dt$$$$=\mathrm{\Psi }\left(1\right)+{\int }_0^{\mathrm{\infty }}\frac{t^2{e}^{-t}(1-e^{-t})+{te}^{-2t}(1-e^{-t})-e^{-t}(1+e^{-2t}-2e^{-t})}{t^2(1-e^{-t})}dt$$$$=\mathrm{\Psi }\left(1\right)+{\int }_0^{\mathrm{\infty }}\frac{t^2{e}^{-t}+{te}^{-2t}-e^{-t}(1-e^{-t})}{t^2}dt=\mathrm{\Psi }\left(1\right)+T$$$$T={\int }_0^{\mathrm{\infty }}{e}^{-t}dt+{\int }_0^{\mathrm{\infty }}\frac{{te}^{-2t}-e^{-t}+e^{-2t}}{t^2}{e}^{-st}dt=1+w\left(s\right)$$$$T=w\left(0\right)+1$$$$w^{\prime }\left(s\right)={\int }_0^{\mathrm{\infty }}-e^{-(2+s)t}dt+{\int }_0^{\mathrm{\infty }}\frac{e^{-t(1+s)}-e^{-t(2+s)}}{t}dt$$$$w^{\prime }\left(s\right)=-\frac{1}{2+s}-{\int }_0^{\mathrm{\infty }}{\int }_{2+s}^{1+s}{e}^{-zt}dzdt$$$$w^{\prime }\left(s\right)=-\frac{1}{2+s}-{\int }_{2+s}^{1+s}{\int }_0^{\mathrm{\infty }}{e}^{-zt}dtdz$$$$=-\frac{1}{2+s}+ln\left(\frac{2+s}{1+s}\right)$$$$w\left(s\right)=-ln(2+s)+(s+2)ln(s+2)-(1+s)ln(1+s)+c$$$$w\left(s\right)=(s+1)ln\left(\frac{s+2}{s+1}\right)+c$$$$(s+1)ln(1+\frac{1}{1+s})+c$$$$\underset{s\to \mathrm{\infty }}{\lim }w\left(s\right)=0\Rightarrow \underset{s\to \mathrm{\infty }}{\lim }(s+1)ln(1+\frac{1}{1+s})+c=0$$$$\Rightarrow c=-1$$$$\Rightarrow w\left(0\right)=ln(1+1)-1=ln\left(2\right)-1$$$$T=1+w\left(0\right)=ln\left(2\right)$$$$S=\mathrm{\Psi }\left(1\right)+T=-\gamma +ln\left(2\right)...$$$$$$

Commented by mnjuly1970 last updated on 18/Nov/20

$$bravobravo$$$$sirmindspower.$$$$extraordinarymyfriend.{\left(good\right)}^{\mathrm{\infty }}$$$$$$

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